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erma4kov [3.2K]
2 years ago
10

The mass of the nails and copper sulfate solution used in this investigation was 250g The mass of the copper and tran sulfate so

lution produced in this
nvestigation would be
250g
Chemistry
2 answers:
likoan [24]2 years ago
5 0
He it would because 250g is the answer is to 256g
Sati [7]2 years ago
3 0

Answer:

He it would because 250g is the answer is to 256g

Explanation:

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A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 ml of 0.106 m naoh. calculat
Jet001 [13]
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions.  So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3. 
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole.  So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
6 0
3 years ago
A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur
dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

So, mass of O_{2} must be withdrawn = (32\times 0.40)g=12.8g

7 0
2 years ago
I will mark brainliest
Vikentia [17]

It's an example of a molecule

6 0
3 years ago
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Please help ASAP. No one could not help me with this question.
Tanya [424]

Answer:

sorry

Explanation:

I don't know the answer this is really confusing but I am really sorry you have to do this.

3 0
2 years ago
Which statement best describes why specific heat capacity is often more useful than heat capacity for scientists when comparing
Misha Larkins [42]

The correct option is this: SPECIFIC HEAT CAPACITY IS AN INTENSIVE PROPERTY AND DOES NOT DEPEND ON SAMPLE SIZE.

Generally, all the properties of matters can be divided into two classes, these are intensive and extensive properties. Intensive properties are those properties that are not determined by the quantity of the material that is present or available. Examples of intensive properties are colour, density and specific heat capacity. For instance, whether you have a bucket of water or a cup of water, the quantity does not matter, the colour of water will always remain the same. Extensive properties in contrast, are those properties that depend on the quantity of material that is available. Examples are mass, heat capacity and volume.

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3 years ago
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